MHB How to derive $P(PH)$ without using a joint distribution table?

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To derive $P(PH)$ without a joint distribution table, the sum rule and product rule can be applied. Using the given probabilities, $P(PH)$ can be calculated as $P(PH | H)P(H) + P(PH | \lnot H)P(\lnot H)$. Substituting the values, $P(PH | H) = 0.8$, $P(PH | \lnot H) = 0.3$, $P(H) = 0.1$, and $P(\lnot H) = 0.9$, results in $P(PH) = 0.8 \times 0.1 + 0.3 \times 0.9$. This calculation yields a final result for $P(PH)$.
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Given $P(PH | H) = 0.8$ and $P(PH | \lnot H) = 0.3 $ and $P(H) = 0.1$ how can I derive $P(PH)$ without resorting to a joint distribution table?
 
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tmt said:
Given $P(PH | H) = 0.8$ and $P(PH | \lnot H) = 0.3 $ and $P(H) = 0.1$ how can I derive $P(PH)$ without resorting to a joint distribution table?

Hi tmt,

We can use that (applying sum rule and general product rule):
$$P(A) = P((A\land B)\lor (A\land \lnot B)) = P(A\land B) + P(A\land\lnot B) = P(A\mid B)P(B) + P(A\mid \lnot B)P(\lnot B)$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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